In the reaction
`BrO^(-3)(aq) + 5Br^(-) (aq) + 6H^(+) rarr 3Br_(2)(1) + 3H_(2)O(1)`
The rate of appearance of bromine `(Br_(2))` is related to rate of disapperance of bromide ions as folllwoing :

Options

A. `(d[Br_(2)])/(dt) = -(5)/(3) (d[Br^(-)])/(dt)`
B. `(d[Br_(2)])/(dt)= (5)/(3) (d[Br^(-)])/(dt)`
C. `(d[Br_(2)])= (3)/(5) (d[Br^(-)])/(dt)`
D. `(d[Br_(2)])/(dt) = -(3)/(5) (d[Br^(-)])/(dt)`

Answer

D

Explanation

For the reaction
`BrO_(3)^(-)(aq)+5Br^(-)(aq)+6H^(+)(aq)rarr3Br_(2)(l)+3H_(2)O(l)`
We have
Rate `=-(d[BrO_(3)^(-)])/(dt)=-(d[Br^(-)])/(5dt)=-(d[H^(+)])/(6dt)`
`=(d[Br_(2)])/(3dt)=-(d[H_(2)O])/(3dt)`
Thus
`(d[Br_(2)])/(dt)`=-(3)/(5)(d[Br^(-)])/(dt)`